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Math Help - A non-parametric equation from a parametrization of a surface (sphere?) help please!!

  1. #1
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    A non-parametric equation from a parametrization of a surface (sphere?) help please!!

    Hi, please help I am stuck on this one. I think the surface is that of a sphere or something like that but I cannot work out the above/below equation.

    r(u,v) = (cos u sin v, cos u cos v, sin u)

    I know that you guys dont like to just give the answers without the poster even trying, so here is what I have so far:

    If the problem is: r(u,v) =(cos u sin v, cos u cos v, sinu) as above, then the non-parametric equation looks like the form:

    F(x,y,z) = C where F is a function of the x,y,z coordinates defining the sphere and C is 1 so that it is completely spherical? i think... like x^2, y^2, z^2 = 1

    Can anyone help make this a non-parametric equation?
    edit. Actually x^2+y^2+z^2 =1 is right isnt it? But what is the algebra to derive this?

    Cheers!
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  2. #2
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    Quote Originally Posted by maxmadx View Post
    Hi, please help I am stuck on this one. I think the surface is that of a sphere or something like that but I cannot work out the above/below equation.

    r(u,v) = (cos u sin v, cos u cos v, sin u)

    I know that you guys dont like to just give the answers without the poster even trying, so here is what I have so far:

    If the problem is: r(u,v) =(cos u sin v, cos u cos v, sinu) as above, then the non-parametric equation looks like the form:

    F(x,y,z) = C where F is a function of the x,y,z coordinates defining the sphere and C is 1 so that it is completely spherical? i think... like x^2, y^2, z^2 = 1

    Can anyone help make this a non-parametric equation?
    edit. Actually x^2+y^2+z^2 =1 is right isnt it? But what is the algebra to derive this?

    Cheers!
    This is a vector equation so you have 3 coordinates:
    You have:
    x = cos(u)~sin(v)
    y = cos(u)~cos(v)
    z = sin(u)

    See what you can do with this:
    x^2 + y^2 + z^2= cos^2(u)~sin^2(v) + cos^2(u)~cos^2(v) + sin^2(u)

    -Dan
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  3. #3
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    thanks for the reply

    ok,

    I know that it needs some algebraic manipulation, but I dont know how to do it with sin/cos. like can you show me the first 1 please?

    I can only get this far:

    x^2= cos^2u sin^2v
    cos^2u=x^2-sin^2V
    u=x^2-sin^2v-cos^2

    y^2= cos^2u cos^2v
    cos^2u=y^2-cos^2v
    u=y^2-cos^2v-cos^2

    z^2= sin^2u
    sin^2u=-z^2
    u=-z^2-sin^2

    arggghhh... dun even know what im doing!
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  4. #4
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    ok wait a min...

    Assuming its a unit sphere (This assumption seems wrong to lead to making a solution.. but anyway..)

    For a unit sphere: (x, y, z) = r(u,v)
    = (u,v, root of 1-u^2-v^2)

    if u=x, and v=y, and z=1-x^2-y^2
    simplified to: z^2=1-x^2-y^2

    then x^2+y^2+z^2=1 And thats the solution according to my answer sheet.
    However, how do I answer the problem correctly...given that x,y,z are defined through cos/sin vectors.

    There is an interm step x^2 + y^2 = cos^2u in the solution sheet i got, but I dont understand how it got to that point
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